This question is an exact duplicate of:
A well-known old problem is to ﬁnd a polynomial function, $f(x)$, such that, for each positive integer $n = 1,2,3,...,$ the result $f(n)$ is prime. Notice that if such a function exists, then one would have an alternate proof of the inﬁnitude of primes, since there must be inﬁnitely diﬀerent outputs among the numbers $f(1),f(2),f(3),....$ (a) Prove the last statement. That is, assume that an nth degree polynomial only generated ﬁnitely many diﬀerent values among $f(1),f(2),f(3)$,... and derive a contradiction.
Assume a finite number of primes. Then we say that there are $n$ primes, and we can list them in order: let $2=p_1<p_2<p_n$ be all primes. Now we define an integer $N=1+p_1*p_2*....*p_2$. Since $N>p_n$ and $p_n$ is the larget prime, $N$ is not prime. However, by the fundamental theorem of arithmetic (using the lemma $N>1)N$ must have a prime factor. This means of the primes in our list must divide $N$, In other words there exist an integer $i$ with $1\le i \le n$ such that $p_i$ divides $N$. Since $p_i$ divides both $N$ and the products of all the primes; it must also divide $N-p_1*p_2*...*p_n=1$. Since $p_i\ge 2$ it is impossible that $p_i$ divides one. Hence, a contradiction.
I wrote this up and presented it to my professor he stated that it does not address the point of polynomial functions. I don't think i am seeing this question correctly. Can anyone assist in the proof to address the problem.